Minimum number of Apples to be collected from trees to guarantee M red apples

There are different kinds of apple trees in the four directions (East, West, North, South), which may grow both red and green apples such that each tree grows exactly K apples, in the following manner:

• N – number of trees to the north does not have red apples.
• S – number of trees to the south does not have green apples.
• W – number of trees in the west has some red apples.
• E – number of trees in the east have some green apples.

However, the colors of apples cannot be distinguished outside the house. So, the task is to find the minimum number of apples to be collected from the trees to guarantee M red apples. If it is not possible, print -1.

Examples:

Input: M = 10, K = 15, N = 0, S = 1, W = 0, E = 0
Output: 10
Explanation: It simply gets 10 apples from the 1st south tree

Input: M = 10, K = 15, N = 3, S = 0, W = 1, E = 0
Output: -1
Explanation: There are no red apples in the South, North and East. But in the West there are atleast 1 red apple and total tree is 1, So, total no. of guaranteed red apple is 1 * 1 = 1 which is less than M.

Approach:  Every apple in the south ensures that it is red. So first, take an apple from the south. In the East and West, there is at least 1 red apple in each tree. That’s why for guaranteed it is considered that there is only 1 red apple on each tree in the east and west. For the north there is no red apple, so, neglect that. Follow the steps below to solve the problem:

• If M is less than equal to S*K then print M.
• Else if M is less than equal to S*K+E+W then print S*K + (M-S*K) * K
• Else print -1.

Below is the implementation of the above approach: 

C++

// C++ program for the above approach
#include<bits/stdc++.h>
using namespace std;

// Function to minimum no. of apples
int minApples(int M,int K,int N,int S,int W,int E){
    // If we get all required apple
    // from South
    if(M <= S * K)
        return M;

    // If we required trees at
    // East and West
    else if(M <= S * K + E + W)
        return S * K + (M-S * K) * K;

    // If we doesn't have enough
    // red apples
    else
        return -1;

}

// Driver Code
int main(){

    // No. of red apple for gift
    int M = 10;

    // No. of red apple in each tree
    int K = 15;

    // No. of tree in North
    int N = 0;

    // No. of tree in South
    int S = 1;

    // No. of tree in West
    int W = 0;

    // No. of tree in East
    int E = 0;

    // Function Call
    int ans = minApples(M,K,N,S,W,E);
    cout<<ans<<endl;

}

// This code is contributed by ipg2016107.

Java

// Java program for the above approach
import java.io.*;
class GFG {

// Function to minimum no. of apples
static int minApples(int M,int K,int N,int S,int W,int E)
{

    // If we get all required apple
    // from South
    if(M <= S * K)
        return M;

    // If we required trees at
    // East and West
    else if(M <= S * K + E + W)
        return S * K + (M-S * K) * K;

    // If we doesn't have enough
    // red apples
    else
        return -1;
}

// Driver code
public static void main(String[] args)
{
    // No. of red apple for gift
    int M = 10;

    // No. of red apple in each tree
    int K = 15;

    // No. of tree in North
    int N = 0;

    // No. of tree in South
    int S = 1;

    // No. of tree in West
    int W = 0;

    // No. of tree in East
    int E = 0;

    // Function Call
    int ans = minApples(M,K,N,S,W,E);
    System.out.println(ans);
}
}

// This code is contributed by code_hunt.

Python3

# Python program for the above approach


# Function to minimum no. of apples
def minApples():

    # If we get all required apple
    # from South
    if M <= S * K:
        return M

    # If we required trees at
    # East and West
    elif M <= S * K + E + W:
        return S * K + (M-S * K) * K

    # If we doesn't have enough
    # red apples
    else:
        return -1


# Driver Code
if __name__ == "__main__":

    # No. of red apple for gift
    M = 10

    # No. of red apple in each tree
    K = 15

    # No. of tree in North
    N = 0

    # No. of tree in South
    S = 1

    # No. of tree in West
    W = 0

    # No. of tree in East
    E = 0

    # Function Call
    ans = minApples()
    print(ans)

C#

// C# program for the above approach
using System;

class GFG{

// Function to minimum no. of apples
static int minApples(int M, int K, int N,
                     int S, int W, int E)
{

    // If we get all required apple
    // from South
    if (M <= S * K)
        return M;

    // If we required trees at
    // East and West
    else if (M <= S * K + E + W)
        return S * K + (M - S * K) * K;

    // If we doesn't have enough
    // red apples
    else
        return -1;
}

// Driver code
public static void Main(String[] args)
{

    // No. of red apple for gift
    int M = 10;

    // No. of red apple in each tree
    int K = 15;

    // No. of tree in North
    int N = 0;

    // No. of tree in South
    int S = 1;

    // No. of tree in West
    int W = 0;

    // No. of tree in East
    int E = 0;

    // Function Call
    int ans = minApples(M, K, N, S, W, E);
    Console.Write(ans);
}
}

// This code is contributed by shivanisinghss2110

Javascript

<script>

      // JavaScript program for the above approach;

      // Function to minimum no. of apples
      function minApples() {

          // If we get all required apple
          // from South
          if (M <= S * K)
              return M;

          // If we required trees at
          // East and West
          else if (M <= S * K + E + W)
              return S * K + (M - S * K) * K;

          // If we doesn't have enough
          // red apples
          else
              return -1;
      }

      // Driver Code

      // No. of red apple for gift
      M = 10

      // No. of red apple in each tree
      K = 15

      // No. of tree in North
      N = 0

      // No. of tree in South
      S = 1

      // No. of tree in West
      W = 0

      // No. of tree in East
      E = 0

      // Function Call
      ans = minApples()
      document.write(ans);

 // This code is contributed by Potta Lokesh
  </script>

Output

10

Time Complexity: O(1) // since no loop is used the algorithm takes constant space to execute
Auxiliary Space: O(1) // since no extra array is used the solution takes up constant space.